Problem: Let $R$ be the region in the second and third quadrants enclosed by the polar curve $r(\theta)=\theta\cdot\cos^2(\theta)$, as shown in the graph. $y$ $x$ $R$ $ 1$ $ 1$ Which integral represents the area of $R$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $ \int_{0}^{\pi}\dfrac{1}{2}\cdot\theta^2\cdot\cos^4(\theta)\,d\theta$ (Choice B) B $ \int_{0}^{\pi}\theta^2\cdot\cos^4(\theta)\,d\theta$ (Choice C) C $ \int_{\scriptsize\dfrac{\pi}{2}}^{\scriptsize\dfrac{3\pi}{2}}\theta^2\cdot\cos^4(\theta)\,d\theta$ (Choice D) D $ \int_{\scriptsize\dfrac{\pi}{2}}^{\scriptsize\dfrac{3\pi}{2}}\dfrac{1}{2}\cdot\theta^2\cdot\cos^4(\theta)\,d\theta$
Answer: This is the formula for the area enclosed by a polar curve $r(\theta)$ between $\theta=\alpha$ and $\theta=\beta$ : $ \int_{\alpha}^{\beta}\dfrac{1}{2}\left(r(\theta)\right)^{2}d\theta$ We know $r(\theta)$ but we still need to figure out $\alpha$ and $\beta$. The part of the curve that encloses $R$ starts at a point where $r(\theta)=0$ and ends at the next point where $r(\theta)=0$. The first non-negative $\theta$ -values for which $r(\theta)=0$ are $0$, $\dfrac{\pi}{2}$, and $\dfrac{3\pi}{2}$. So $\alpha$ must be $\dfrac{\pi}{2}$ and $\beta$ must be $\dfrac{3\pi}{2}$. Let's plug ${r(\theta)=\theta\cdot\cos^2(\theta)}$, ${\alpha=\dfrac{\pi}{2}}$, and ${\beta=\dfrac{3\pi}{2}}$ into the formula and expand the parentheses: $\begin{aligned} &\phantom{=} \int_{\alpha}^{\beta}\dfrac{1}{2}\left({r(\theta)}\right)^{2}d\theta \\\\ &= \int_{{\scriptsize\dfrac{\pi}{2}}}^{{\scriptsize\dfrac{3\pi}{2}}}\dfrac{1}{2}\left({\theta\cdot\cos^2(\theta)}\right)^{2}d\theta \\\\ &= \int_{\scriptsize\dfrac{\pi}{2}}^{\scriptsize\dfrac{3\pi}{2}}\dfrac{1}{2}\cdot\theta^2\cdot\cos^4(\theta)\,d\theta \end{aligned}$ In conclusion, this integral represents the area of region $R$ : $ \int_{\scriptsize\dfrac{\pi}{2}}^{\scriptsize\dfrac{3\pi}{2}}\dfrac{1}{2}\cdot\theta^2\cdot\cos^4(\theta)\,d\theta$